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QJEEAdv22Mad127: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Mad127 Solution: Let be nonzero real numbers that are, respectively, the and…

Let p,q,rp, q, r be nonzero real numbers that are, respectively, the 10th ,100th 10^{\text {th }}, 100^{\text {th }} and 1000th 1000^{\text {th }} terms of a harmonic progression. Consider the system of linear equations

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List-I List-II
(I) If qr=10\frac{q}{r}=10 , then the system of linear equations has (P) x=0,y=109,z=−19x=0, \quad y=\frac{10}{9}, z=-\frac{1}{9} as a solution
(II) If pr≠100\frac{p}{r} \neq 100 , then the system of linear equations has (Q) x=109,y=−19,z=0x=\frac{10}{9}, y=-\frac{1}{9}, z=0 as a solution
(III) If pq≠10\frac{p}{q} \neq 10 , then the system of linear equations has (R) infinitely many solutions
(IV) If pq=10\frac{p}{q}=10 , then the system of linear equations has (S) no solution
(T) at least one solution

The correct option is:

Multiple Choice Options

  • A (I) →\rightarrow (T); (II) →\rightarrow (R); (III) →\rightarrow (S); (IV) →\rightarrow (T)
  • B (I) →\rightarrow (Q); (II) →\rightarrow (S); (III) →\rightarrow (S); (IV) →\rightarrow (R)
  • C (I) →(Q)\rightarrow(\mathrm{Q}) ; (II) →\rightarrow (R); (III) →(P)\rightarrow(\mathrm{P}) ; (IV) →\rightarrow (R)
  • D (I) →\rightarrow (T); (II) →\rightarrow (S); (III) →\rightarrow (P); (IV) →\rightarrow (T)

Step-by-Step Verified Solution

Given:

x+y+z=1..........(1)10x+100y+1000z=0.........(2)qrx+pry+pqz=0............(3)\begin{gathered}x+y+z=1 ..........(1)\\\\10 x+100 y+1000 z=0 .........(2)\\\\q r x+p r y+p q z=0 ............(3)\end{gathered}

Now equation (3) can be re-written as

xp=yq=zr=0∵{p,q,r≠0}\frac{x}{p}=\frac{y}{q}=\frac{z}{r}=0 \quad \because\{p, q, r \neq 0\}

Now given p,qp, q and rr are 10th ,100th 10^{\text {th }}, 100^{\text {th }} and 1000th 1000^{\text {th }} term of an. H.P.,

So let p=ya+9d,q=1a+99d,r=1a+999dp=\frac{y}{a+9 d}, q=\frac{1}{a+99 d}, r=\frac{1}{a+999 d}

Now from equation (3)

(a+9d)x+(a+99d)y+(a+999d)z=0(a+9 d) x+(a+99 d) y+(a+999 d) z=0

Now from and (1), (2) and (3) we get

Δamp;=∣1amp;1amp;110amp;100amp;1000a+9damp;a+99damp;a+999d∣=0Δxamp;=∣1amp;1amp;10amp;100amp;10000amp;a+99damp;a+999d∣=900(d−a)Δyamp;=∣10amp;0amp;10001+9damp;0amp;a+999d∣=990(a−d)Δzamp;=∣1amp;1amp;11amp;1amp;110amp;100amp;0a+9damp;a+99damp;0∣=90(d−a)\begin{aligned}\Delta &amp; =\left|\begin{array}{ccc}1 &amp; 1 &amp; 1 \\10 &amp; 100 &amp; 1000 \\a+9 d &amp; a+99 d &amp; a+999 d\end{array}\right|=0 \\\\\Delta x &amp; =\left|\begin{array}{ccc}1 &amp; 1 &amp; 1 \\0 &amp; 100 &amp; 1000 \\0 &amp; a+99 d &amp; a+999 d\end{array}\right|=900(d-a) \\\\\Delta y &amp; =\left|\begin{array}{ccc}10 &amp; 0 &amp; 1000 \\1+9 d &amp; 0 &amp; a+999 d\end{array}\right|=990(a-d) \\\\\Delta z &amp; =\left|\begin{array}{ccc}1 &amp; 1 &amp; 1 \\1 &amp; 1 &amp; 1 \\10 &amp; 100 &amp; 0 \\a+9 d &amp; a+99 d &amp; 0\end{array}\right|=90(d-a)\end{aligned}

(I) If qr=10⇒a=d\frac{q}{r}=10 \Rightarrow a=d

Δ=Δx=Δy=Δz=0\Delta=\Delta_x=\Delta_y=\Delta_z=0

And equation (1) and equation (2) represents non-parallel plane equation (2) and equation (3) represents same plane

⇒\Rightarrow Infinitely many solutions.

Now finding solution by taking z=λz=\lambda then

amp;x+y=1−λamp;x+10y=−100λamp;x=109+100λamp;y=−19−11λamp;⇒(x,y,z)∈(109+10λ,−19−11λ,λ)\begin{aligned}&amp; x+y =1-\lambda \\\\&amp; x+10 y =-100 \lambda \\\\&amp; x =\frac{10}{9}+100 \lambda \\\\&amp; y =\frac{-1}{9}-11 \lambda \\\\&amp; \Rightarrow (x, y, z) \in\left(\frac{10}{9}+10 \lambda, \frac{-1}{9}-11 \lambda, \lambda\right)\end{aligned}

So P\mathrm{P} is not valid for any value of λ→Q\lambda \rightarrow \mathrm{Q}

(II) pr≠100⇒a≠d\frac{p}{r} \neq 100 \Rightarrow a \neq d

Δ=0 and Δx′Δy′Δz≠0\Delta=0 \,\text{and}\, \Delta_{x^{\prime}} \Delta_{y^{\prime}} \Delta_z \neq 0

So no solution.

 II →S\text { II } \rightarrow \mathrm{S}

(III) If pq≠10⇒a≠d then Δz≠0\frac{p}{q} \neq 10 \Rightarrow a \neq d \text { then } \Delta_z \neq 0

So no solution.

 III →S\text { III } \rightarrow \mathrm{S}

(IV) If ⇒a=d then Δz=0⇒Δx=Δy=0\Rightarrow a=d \text { then } \Delta_z=0 \Rightarrow \Delta_x=\Delta_y=0

So infinitely many solutions.

IV→R\mathrm{IV} \rightarrow \mathrm{R}

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